हिंदी

Examine the differentiability of f, where f is defined byf(x) = ,if,if{x2sin 1x, if x≠00,if x=0 at x = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

Examine the differentiability of f, where f is defined by
f(x) = `{{:(x^2 sin  1/x",",  "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0

योग
Advertisements

उत्तर

Given that, f(x) = `{{:(x^2 sin  1/x",",  "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0

For differentiability we know that:

Lf'(c) = Rf'(c)

∴ Lf'(0) = `lim_("h" -> 0)  ("f"(0 - "h") - "f"(0))/(-"h")`

= `lim_("h" -> 0) ((0 - "h")^2 sin  1/((0 - "h")) - 0)/(-"h")`

= `("h"^2 sin  (- 1/"h"))/(-"h")`

= `"h"* sin (1/"h")`

= `0 xx [-1 ≤ sin  (1/"h") ≤ 1]`

= 0

Rf'(0) = `lim_("h" -> 0)  ("f"(0 + "h") - "f"(0))/"h"`

= `lim_("h" -> 0)  ((0 + "h")^2 sin (1/(0 + "h") - 0))/"h"`

= `lim_("h" -> 0) ("h"^2 sin (1/"h"))/"h"`

= `lim_("h" -> 0) "h" * sin (1/"h")`

= `0 xx [-1 ≤ sin (1/"h") ≤ 1]`

= 0

So, Lf'(0) = Rf'(0) = 0

Hence, f(x) is differentiable at x = 0.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ १०९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 21 | पृष्ठ १०९

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

Examine the following function for continuity:

f(x) = `(x^2 - 25)/(x + 5)`, x ≠ −5


Examine the following function for continuity:

f(x) = |x – 5|


Discuss the continuity of the function f, where f is defined by:

f(x) = `{(3", if"  0 <= x <= 1),(4", if"  1 < x < 3),(5", if"  3 <= x <= 10):}`


Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - 1 \right|}{x - 1}, for & x \neq 1 \\ 2 , for & x = 1\end{cases}at x = 1\]

Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{2\left| x \right| + x^2}{x}, & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]

Discuss the continuity of the function f(x) at the point x = 0, where  \[f\left( x \right) = \begin{cases}x, x > 0 \\ 1, x = 0 \\ - x, x < 0\end{cases}\]

 


Determine the value of the constant k so that the function 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{x^2 - 3x + 2}{x - 1}, if & x \neq 1 \\ k , if & x = 1\end{array}\text{is continuous at x} = 1 \right.\] 


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  \[f\left( x \right) = \begin{cases}\frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k , & x = 5\end{cases}\]at x = 5


Let\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if }  x < \frac{\pi}{2} \\ a , & \text{ if }  x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if }  x > \frac{\pi}{2}\end{array} . \right.\] ]If f(x) is continuous at x = \[\frac{\pi}{2}\] , find a and b.

 

Discuss the continuity of the function  \[f\left( x \right) = \begin{cases}2x - 1 , & \text { if }  x < 2 \\ \frac{3x}{2} , & \text{ if  } x \geq 2\end{cases}\]


Find all point of discontinuity of the function 

\[f\left( t \right) = \frac{1}{t^2 + t - 2}, \text{ where }  t = \frac{1}{x - 1}\]

If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\]  , then

 


The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & 0 \leq x \leq 1\end{cases}\]is continuous in the interval [−1, 1], then p is equal to

 


The function  \[f\left( x \right) = \frac{x^3 + x^2 - 16x + 20}{x - 2}\] is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as ______.


If  \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (ab) for which f (x) cannot be continuous at x = 1, is

 


If  \[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x} , & x < \frac{\pi}{2} \\ a , & x = \frac{\pi}{2} \\ \frac{b\left( 1 - \sin x \right)}{\left( \pi - 2x \right)^2}, & x > \frac{\pi}{2}\end{cases}\]. Then, f (x) is continuous at  \[x = \frac{\pi}{2}\], if

 


Show that f(x) = |x − 2| is continuous but not differentiable at x = 2. 


Show that f(x) = x1/3 is not differentiable at x = 0.


If \[f\left( x \right) = \begin{cases}a x^2 - b, & \text { if }\left| x \right| < 1 \\ \frac{1}{\left| x \right|} , & \text { if }\left| x \right| \geq 1\end{cases}\]  is differentiable at x = 1, find a, b.


Discuss the continuity and differentiability of f (x) = |log |x||.


Write the points of non-differentiability of 

\[f \left( x \right) = \left| \log \left| x \right| \right| .\]

The function f (x) = sin−1 (cos x) is


Examine the continuity off at x = 1, if

f (x) = 5x - 3 , for 0 ≤ x ≤ 1

       = x2 + 1 , for 1 ≤ x ≤ 2


 If the function f (x) = `(15^x - 3^x - 5^x + 1)/(x tanx)`,  x ≠ 0 is continuous at x = 0 , then find f(0).


Find `dy/dx if y = tan^-1 ((6x)/[ 1 - 5x^2])`


If y = ( sin x )x , Find `dy/dx`


Discuss the continuity of function f at x = 0.
Where f(X) = `[ [sqrt ( 4 + x ) - 2 ]/ ( 3x )]`, For x ≠ 0
                  = `1/12`,                      For x = 0


Examine the continuity of the followin function : 

  `{:(,f(x),=x^2cos(1/x),",","for "x!=0),(,,=0,",","for "x=0):}}" at "x=0`   


If f(x) = `(sqrt(2) cos x - 1)/(cot x - 1), x ≠ pi/4` find the value of `"f"(pi/4)`  so that f (x) becomes continuous at x = `pi/4`


Let f(x) = `{{:((1 - cos 4x)/x^2",",  "if"  x < 0),("a"",",  "if"  x = 0),(sqrt(x)/(sqrt(16) + sqrt(x) - 4)",", "if"  x > 0):}`. For what value of a, f is continuous at x = 0?


The function f(x) = |x| + |x – 1| is ______.


The set of points where the functions f given by f(x) = |x – 3| cosx is differentiable is ______.


f(x) = |x| + |x − 1| at x = 1


Find the values of a and b such that the function f defined by
f(x) = `{{:((x - 4)/(|x - 4|) + "a"",",  "if"  x < 4),("a" + "b"",",  "if"  x = 4),((x - 4)/(|x - 4|) + "b"",", "if"  x > 4):}`
is a continuous function at x = 4.


A function f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f′(0) = 2. Prove that f′(x) = 2f(x).


The set of points where the function f given by f(x) = |2x − 1| sinx is differentiable is ______.


If f(x) = `{{:("m"x + 1",",  "if"  x ≤ pi/2),(sin x + "n"",",  "If"  x > pi/2):}`, is continuous at x = `pi/2`, then ______.


If f is continuous on its domain D, then |f| is also continuous on D.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×